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A lion and its tamer

If I remember well, if the tamer runs on the perimeter, the lion catches the tamer.
If the lion follows the radius of the tamer, his trajectory is a 1/2 circle and not a spiral. He catches the tamer in a finite time.

The good news is that using equations for a more efficient approach strategy by the lion as the trainer runs around the perimeter doesn't require trig functions. The bad news is that I nonetheless end up with a differential equation that I can't see how to solve.

Spoiler for redo of the lion's approach path

If the tamer moves a distance dtheta along the perimeter, the "lateral" distance that the lion has to run to stay on a line connecting the center and the tamer is r*dtheta just like before. But this time, instead of having the lion run along an inner circle's path for r*dtheta and then outward for (1-r)dtheta, I'll ask how far outward the lion can go by running along a skewed arc of total length dtheta with a constantly increasing r as the lion goes from the start to the finish of this distance.

Because r is changing as the lion is running we can't just use r*dtheta as the total radial distance that the lion moves to stay on a line connecting the center and tamer; we have to account for the fact that the circumference of the circle he's following grows larger as r increases. But since circumference is linearly proportional to radius, we can just use the average radius as the lion is moving outward: (r + (r+dr)) / 2 = r + dr/2. So the "lateral" distance the lion moves is dtheta*(r + dr/2). The "outward" distance the lion moves is dr. The "total" distance the lion moves is dtheta (the same total distance that the tamer moved). Then use the pythagorean theorem to find out how large dr can be in relation to dtheta for any given r.

dtheta2 = [dtheta*(r + dr/2)]2 + dr2

Integrating that function with the initial conditions of r=0 and theta=0 should let you solve for how large theta is when r reaches 1, and thus how far the tamer runs before he gets eaten. I'm not sure if I would have been able to solve that differential equation back when I was taking calculus classes years ago, but I sure can't now.

But I can show that the lion is able to catch the tamer, albeit with a much less optimal approach.

Spoiler for

With the first method I posted, the lion can get very close to the tamer but not quite catch him. Let the lion use the first strategy to reach a distance r from the center on a line toward the tamer. Then the lion's strategy will change, and the lion will now run perpendicular to the line connecting the center and the tamer. If the tamer runs along the perimeter for a very small distance dx, then the lion will run a distance r*dx perpendicular to the radius. After running for r*dx perpendicular to the radius, the lion's new distance from the center will be sqrt(r2 + (r*dx)2) = sqrt(r2*(1+dx2)) = r*sqrt(1+dx2).

Again, actually turning that into a differential equation and solving it is beyond me, but the key point to notice is that r is always being multiplied by sqrt(1+dx2) and will be continuously increasing at a rate that does not approach zero, which is enough to show that the lion will catch the tamer. (Although it's funny that with this strategy the lion can't make it out of the center: the increase in r is r*sqrt(1+dx2), so zero when r=0, hence why I made the lion use the first strategy to get out and then switch to this second strategy to reach the tamer.)

I thought lions could pounce. But even if that is not allowed, the lion still must eat.

Spoiler for Change of plan for the hungry lion

Maybe, cutting the prey off, like you do it with zebra, will work for this hunt as well.

After running directly towards the tamer for a bit (half a unit, or so,) the lion could just move directly to the cage wall. Whereupon, he will project the rendezvous point at the perimeter (assuming the trainer continues to walk along the wall) and sprint directly to that point in a straight line of a chord.
E.g. if initially the central angle between the lion on the perimeter and the man on the perimeter is a, and the angle between lion’s starting position and projected meeting point is x, the lion could solve an equation to project the meeting place (all the while continuing to run):
(x-a) = x*(2-2*cos x)1/2. (Using the formula for the chord length and taking into account the unit radius.)
If the quarry changes his path, the lion can recalculate meeting point for the new trajectory of the prey, head to it in a straight line and have his lunch even sooner.

I thought lions could pounce. But even if that is not allowed, the lion still must eat.[/size]

Spoiler for Change of plan for the hungry lion

[/size]Maybe, cutting the prey off, like you do it with zebra, will work for this hunt as well.[/size]

After running directly towards the tamer for a bit (half a unit, or so,) the lion could just move directly to the cage wall. Whereupon, he will project the rendezvous point at the perimeter (assuming the trainer continues to walk along the wall) and sprint directly to that point in a straight line of a chord.
E.g. if initially the central angle between the lion on the perimeter and the man on the perimeter is a, and the angle between lion’s starting position and projected meeting point is x, the lion could solve an equation to project the meeting place (all the while continuing to run):
(x-a) = x*(2-2*cos x)1/2. (Using the formula for the chord length and taking into account the unit radius.)
If the quarry changes his path, the lion can recalculate meeting point for the new trajectory of the prey, head to it in a straight line and have his lunch even sooner.

[/size]

Yes the lion can pounce, I.e. Feet can leave the ground. His top speed is still 1 radius/second.

If the tamer's minutes, or seconds, are measured, how long does he have, with starting locations at center (lion) and perimeter (tamer)?

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A leader's credentials do not hang on his wall. They are written by the lives of those he has led.

If I understand correctly, the Zeno argument says that (going in the reverse direction) when you are at the perimeter the change in the lion's distance from the center over time equals zero so there is no way to get out from the perimeter.

Would this be the same as considering the function y = x2 and saying that dy/dx at x=y=0 is zero, so there is no way for the y value to increase and ever leave the x-axis? Since y is not changing and x is the square root of y, would that also imply that x cannot change and so the function can never even leave the origin at all?

If I understand correctly, the Zeno argument says that (going in the reverse direction) when you are at the perimeter the change in the lion's distance from the center over time equals zero so there is no way to get out from the perimeter.

Would this be the same as considering the function y = x2 and saying that dy/dx at x=y=0 is zero, so there is no way for the y value to increase and ever leave the x-axis? Since y is not changing and x is the square root of y, would that also imply that x cannot change and so the function can never even leave the origin at all?

Yes, sort of.

I don't have the equations and I expected the lion would approach the edge of the cage asymptotically, never actually reaching it. But now I think that's wrong. It is more like a parabola, and the lion's path is simply tangent to the circular edge, touching it at a single, well-defined point. The path has finite length and is retraceable.

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A leader's credentials do not hang on his wall. They are written by the lives of those he has led.

If I remember well, if the tamer runs on the perimeter, the lion catches the tamer.
If the lion follows the radius of the tamer, his trajectory is a 1/2 circle and not a spiral. He catches the tamer in a finite time.

Spoiler for If the tamer is not on the perimeter

Then, he has an escape strategy that works for an infinite time.

Spoiler for Grimbal has the path shape

Starting the lion at (0,0) and the tamer at (1,0) at same speed.

The paths coincide after the tamer has gone 1/4 circle, or pi/2 radii, which, at unit speed, takes pi/2 seconds.

The lion's semicircular path has the same length as the quarter circle, having 1/2 the diameter of the latter circle.

The interesting thing about this chase is that the lion runs directly toward the tamer at only one moment in time: at the very start.

At the point of capture they are running in precisely the same direction, the tamer being at the right side of the lion.

A ray from the origin intersects the two paths where they are occupied at the same point in time.

Another way of saying that in order to stay on a common radius, the lion must face increasingly away from the tamer toward the eventual point of capture.

If the lion runs straight toward the tamer, the tamer escapes. Or does he? Suppose they are both on the perimeter, separated by a very small distance. Does the lion catch up? Perhaps. By running straight toward the tamer, the lion decreases his radius slightly, allowing him to reduce the angular separation. Hmm. Subject for a different puzzle perhaps.

Edit: By running straight toward the tamer, the lion traces a path different from the semicircle (a path whose radius is greater at each point in time than the corresponding point on the semicircle.) But, lacking a linear speed advantage, the lion does not want the tamer directly in front of him. Nevertheless, so long as the lion's radius < 1, he has an angular speed advantage, and eventually he captures the tamer. So this does define a new calculation: how much extra time does that chase strategy on the part of the lion give to the tamer?

Summary:

The tamer becomes the lion's lunch, and he has pi/2 seconds to live once the chase is on.

I'm marking the puzzle solved but invite Grimbal to describe an off-perimeter escape strategy for the tamer.

Edited by bonanova, 30 January 2014 - 02:01 PM.Describe another capture path and pose a new calculation

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A leader's credentials do not hang on his wall. They are written by the lives of those he has led.